Force on a proton in electric field #vecE# is given by

#vecF=qvecE#

Acceleration produced is #veca=q/mvecE#

where #q and m# are the proton's charge and mass.

Treating orthogonal velocities independently.

Time taken by proton to travel horizontal distance #d# to just exit out of electric field

#t=d/v# ......(1)

Vertical component of velocity #v_(xt)# attained in this time is found by the following kinematic expression

#v=u+at#

#v_(xt)=q/m|vecE|d/v#

#v_(xt)=(qd)/(mv)|vecE|# .....(2)

We know that angle of the resultant of two vectors #vecP and vecQ# having angle #theta# between them is given by the formula

#tan alpha=(|vecQ|sintheta)/(|vecP|+|vecQ|costheta)#

We get at the exit point

#tan 15^@=(v_(xt)sin90^@)/(v+v_(xt)cos90^@)#

#=>tan 15^@=(v_(xt))/(v)#

Using (2) we get

#tan 15^@=((qd)/(mv)|vecE|)/(v)#

#=>tan 15^@=(qd|vecE|)/(mv^2)#

#=>|vecE|=(mv^2tan 15^@)/(qd)# .....(3)

Taking charge of proton #q=1.602xx10^-19 C#,

mass of proton #m=1.673 xx 10^-27 kg#

#|vecE|=932.7Vm^-1#

-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Alternate method after equation (1).

During time #t# vertical deflection produced to meet the condition is

#"deflection"/(d/2)=tan15^@#

as the line of exit velocity extended backwards meets the axis of velocity #vecv# at mid-point of #d# *i.e*., at a distance #=d/2#

Setting up the kinematic equation for vertical deflection and noting that initial vertical velocity of proton is zero as it enters the uniform electric field region

#s=ut+1/2at^2#

#d/2tan 15^@=1/2(q/m|vecE|)(d/v)^2#

#d tan 15^@=(q/m|vecE|)(d/v)^2#

#=>|vecE|=m/qd tan15^@(v/d)^2#

#=>|vecE|=(mv^2tan15^@)/(qd)#

which is same as equation (3)